Question

The nucleus of "Lead-208", 208 82 Pb, has 82 protons within a sphere of radius 6.34×10-15 m. Each electric charge has a value of 1.60218 × 10^-19 C. The Coulomb constant is 8.98755 × 109 N · m^2/C^2. Calculate the electric field at the surface of the nucleus. Answer in units of N/C.

Answer 1

Answer:

2.94 × 10²⁰ N/C

Explanation:

Given that:  

The nucleus of "Lead-208 has 82 protons,

with a radius (r) 6.34×10-15 m, &

each electric charge has a value of 1.60218 × 10^-19 C

∴ The formula for calculating an electrical field at the surface of the nucleus is:

 $E=\frac{k*q}{r^2}$  

Substituting our values into the equation above, we have;

 E = $\frac{8.98755*10^8*82(1.60218*10^{-19C)}}{(6.34*10^{-15}_m)^2}$

E = 2.93870499×10²⁰ N/C

E ≅ 2.94 × 10²⁰ N/C